Finding the equation of the intersection

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The discussion focuses on finding the intersection of three planes defined by the equations 2x - 3y + 8z = 7, x - 8y + z = 14, and 5x - 14y + 17z = 28. The user initially derives two contradictory equations, 13y + 6z = -21 and 13y + 6z = -16, leading to the conclusion that no single solution exists for x, y, and z. However, the correct parametric equations are derived as [14/13, -21/13, 0] + t[-61/13, -6/13, 1], demonstrating that a line of intersection exists despite the contradictions in the derived equations.

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Homework Statement


Plane 1:2x-3y+8z=7
Plane 2:x-8y+z=14
Plane 3:5x-14y+17z=28

Homework Equations


N/A


The Attempt at a Solution


I took plane 1 and subtracted (plane 2)x2 to get
13y+6z=-21 (I will refer to this as equation 1)

Then I took (plane 2)x5 and subtracted plane 3 to get
-26y-12z=32
which simplifies to
13y+6z=-16 (I will refer to this as equation 2)

Then I was stumped because the two equations contradict each other and state 0=-5
which leads me to believe that there is no values of x,y,z that can satisfy all planes

although the answers in the book say
"consistent [14/13,-21/13,0] + t[-61/13,-6/13,1]"

I understand how they get the answer I just don't understand why, if equation 1 and 2 do not provide a valid statement why do they continue and find the parametric equations

like let z = t
13y=-6t -21
y=-6/13t -21/13
then substitute y into plane 2 to get
x-8(-6/13z - 21/13) + z=14 and simplifies
x=61/13z +14/13
then z=0+t
which gets the correct parametric equations which I tested by using different values of t and the points do satisfy all planes
I guess the real question I have is why do they chose "equation 1" instead of "equation 2" which would yield a different result if you were to use that to fidn the parametric equations, which I'm assuming the parametric equations don't work and why do you continue after the elimination of equation 1 and 2 which results in 0=-5.
 
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cdot9 said:

Homework Statement


Plane 1:2x-3y+8z=7
Plane 2:x-8y+z=14
Plane 3:5x-14y+17z=28

Homework Equations


N/A


The Attempt at a Solution


I took plane 1 and subtracted (plane 2)x2 to get
13y+6z=-21 (I will refer to this as equation 1)

Then I took (plane 2)x5 and subtracted plane 3 to get
-26y-12z=32
Hello cdot9. Welcome to PF !

5×14 - 28 = 42, not 32.
which simplifies to
13y+6z=-16 (I will refer to this as equation 2)

Then I was stumped because the two equations contradict each other and state 0=-5
which leads me to believe that there is no values of x,y,z that can satisfy all planes

although the answers in the book say
"consistent [14/13,-21/13,0] + t[-61/13,-6/13,1]"

I understand how they get the answer I just don't understand why, if equation 1 and 2 do not provide a valid statement why do they continue and find the parametric equations

like let z = t
13y=-6t -21
y=-6/13t -21/13
then substitute y into plane 2 to get
x-8(-6/13z - 21/13) + z=14 and simplifies
x=61/13z +14/13
then z=0+t
which gets the correct parametric equations which I tested by using different values of t and the points do satisfy all planes
I guess the real question I have is why do they chose "equation 1" instead of "equation 2" which would yield a different result if you were to use that to fidn the parametric equations, which I'm assuming the parametric equations don't work and why do you continue after the elimination of equation 1 and 2 which results in 0=-5.
 

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